# Linear algebra proof question

1. Dec 14, 2008

### transgalactic

the question is in this link?

http://img444.imageshack.us/img444/5041/55025174kd6.gif [Broken]

i dont know how to prove it
i know basic stuff
that Ker T is all the vectors which go to line of zeros in matrix row reduction.
Im T is produced when i traverse the matrix and do a ow reduction
and the remaining(non zero) vectors are Im T
i also know that V=Im T + Ker T

??

Last edited by a moderator: May 3, 2017
2. Dec 14, 2008

### HallsofIvy

Staff Emeritus
The question appears to be asking "Is there a linear transformation T, from R3 to R3 such that ker T is a subset of the inverse of T and such that the kernel of T is spanned by (1, 1, 0). I may be misunderstanding since what I am interpreting as "inverse of T" is shown as "T-1" and if that is so then the problem is trivial: in order to have an inverse, T must be one-to-one- in other words ker T must be {(0,0,0)}. The second questions ask the same about a linear operator S but has kernel of T spanned by (3, 1, 0) and (2, 0, 0). That seems too trivial a question. I may be misunderstanding "T-1" and "S-1".

3. Dec 14, 2008

### transgalactic

sorryy disregard -1 thing
in my language its the word for "AND"
so its first expression and second expression

4. Dec 14, 2008

### HallsofIvy

Staff Emeritus
Well, any linear function from R3 to R3 maps (x, y, z) into (ax+ by+ cz, dx+ ey+ fz, gx+ hy+ iz) so finding a linear function is just a matter of finding 9 numbers. You are told that the kernel is spanned by (3, 1, 0) and (2, 0, 0) so you know that 3a+ b= 0, 3d+ f= 0, 3g+ h= 0, 2a= 0, 2d= 0, and 2g= 0. That should help a lot! Since the kernel is 2 dimensional, you know that the image must be only 1 dimensional: all columns are multiplies of each other. Finally, you want that one dimensional image to be a subspace of the kernel.